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I'm having trouble blending two different types of progressions:

The fourth, eighth and fourteenth terms of an A.P., common difference 0.5, are in geometric progression. Find the first term of the A.P. and the common ratio of the G.P.

In a A.P, $u_n = a + (n-1)d$, so:

$$ u_4 = a + \frac{3}{2}, u_8 = a + \frac{7}{2}, u_{14} = a + \frac{13}{2} $$

but in a geometric series, the difference between $u_n$ and $u_{n+1}$ is $r$:

$$ u_{14} - u_8 = r = \frac{13}{2} - \frac{7}{2} = 3 $$

$$ u_8 - u_4 = r = \frac{7}{2} - \frac{3}{2} = 2 $$

which doesn't make sense, because it looks like $r$ is varying. I also know that the fourth term in the A.P. is offset from the first term by $1.5$ - but I have no means to find $a$. Can somebody point me in the right direction?

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  • $\begingroup$ Hint: We have $u_8^2=u_4u_{14}$. $\endgroup$
    – André Nicolas
    Jul 20, 2014 at 13:59
  • $\begingroup$ @AndréNicolas I don't understand what you mean $\endgroup$
    – hohner
    Jul 20, 2014 at 14:05
  • $\begingroup$ It is equivalent to what Michael wrote. $\endgroup$
    – André Nicolas
    Jul 20, 2014 at 14:09
  • $\begingroup$ @AndréNicolas I still can't figure out how to get $a$ :( $\endgroup$
    – hohner
    Jul 20, 2014 at 14:13
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    $\begingroup$ We have $(a+3/2)(a+13/2)=(a+7/2)^2$. Expand. The $a^2$ terms cancel, so we get a linear equation for $a$. $\endgroup$
    – André Nicolas
    Jul 20, 2014 at 14:15

1 Answer 1

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In a GP, the ratio is $r$. So $$\frac{u_{14}}{u_8} = r = \frac{u_8}{u_4}$$

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